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无穷等差数列{an}各项都是正数,Sn是它的前n项和,若a1+a3+a8=a4^²,则 a5·S4的最大值是
无穷等差数列{an}各项都是正数,Sn是它的前n项和,若a1+a3+a8=a42,则
a5·S4的最大值是______________.
答案是36,但我怎么算都得75/2,
人气:106 ℃ 时间:2019-08-19 05:15:09
解答
a1+a3+a8=3a1+9d=3﹙a1+3d﹚=﹙a1+3d﹚²
a4=a1+3d=3
s4=﹙3+a1﹚*4/2=2﹙3+a1﹚=6+2a1=12-6d
a5=a4+d=3+d
a5*s4
=﹙3+d﹚﹙12-6d﹚
=36+12d-18d-6d²
=36-6d²-6d
=-6﹙d+0.5﹚²+75/2
an>0
a1=3-3d>0
d<1
a1+﹙n-1﹚d>0
-a1<﹙n-1﹚d
d>a1/﹙1-n﹚
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